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Math Help - Question! Binomial Series

  1. #1
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    Question! Binomial Series

    Hi guys I can't do the last part of this question so thanks if anybody could help!

    Find the first four terms, in ascending order, in the expansion of (1-2x)^-2. State the values of x for which this expansion is valid.
    Hence,

    (i) by choosing a suitable value of x, find sum to infinity (starting value r=1) of r/2^(r-1),
    (i) show that [sq root 2 cos (2x+(pi/4))] / (1-2tanx)^2 is estimated to be 1+2x+2x^2, given that x is small enough for its third and higher powers to be ignored.

    Many apologies for the 'messiness'!
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  2. #2
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    Let r be any real number and |y|<1, theny

    (1+y)^r = 1 + \frac{r}{1!}y+\frac{r(r-1)}{2!}y^2+\frac{r(r-1)(r-2)}{3!}y^3+...

    Now, we want to expand (1-2x)^{-2} thus r=-2 and y=-2x. Hence we require that |-2x|=|2x|<1 that is, |x|<\frac{1}{2}.

    Thus the binomial series look like:
    (1-2x)^{-2} = 1 - \frac{-2}{1!}(2y) + \frac{-2(-3)}{2!}(2y)^2 - \frac{-2(-3)(-4)}{3!} (2y)^3+
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